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超几何分布

2022-04-16 20:22 作者:匆匆-cc  | 我要投稿

        先放一段超几何分布的定义。

        一般的,假设一批产品共有%5Ccolor%7Bgray%7D%7BN%7D件,其中有%5Ccolor%7Bgray%7D%7BM%7D件次品,从%5Ccolor%7Bgray%7D%7BN%7D件产品中随机抽取%5Ccolor%7Bgray%7D%7Bn%7D件(不放回),用%5Ccolor%7Bgray%7D%7Bx%7D表示抽取的%5Ccolor%7Bgray%7D%7Bn%7D件产品中的次品数,则

%5Ccolor%7Bgray%7D%7BP_%7B(x%3Dk)%7D%3D%5Cfrac%7BC%5Ek_MC%5E%7Bn-k%7D_%7BN-M%7D%7D%7BC%5En_N%7D%EF%BC%8Ck%3Dm%2Cm%2B1%2Cm%2B2%2C%E2%80%A6%2Cr%7D

        其中%5Ccolor%7Bgray%7D%7Bn%2CN%2CM%5Cin%20N%5E*%2CM%5Cleq%20N%2Cn%5Cleq%20N%2Cm%3D%5Cmax%5C%7B0%2Cn-N%2BM%5C%7D%2Cr%3D%5Cmin%5C%7Bn%2CM%5C%7D%7D

        很烦?

        我们来看通俗版本。

        考虑一批产品,比如说有100件,其中有10件次品。

        我们从中挑选出5件产品。

        那么,如果挑选出了3件正品,2件次品,也就是说要从90件正品中挑选出3件正品,同时从剩余的10件次品挑选出2件次品。

        而这一切,都发生在从100件中挑选出5件产品这一件事中。

        因此,其概率为

P%3D%5Cfrac%7BC%5E%7B3%7D_%7B90%7DC%5E2_%7B10%7D%7D%7BC%5E%7B10%7D_%7B100%7D%7D

        至于其中的定义域问题,主要原因如下:

        仍然考虑一批产品,比如说有100件,其中有10件次品。

        那么,这里面最多挑选出10件次品,即使总共挑选出再多的产品,次品数最多也只有10件。

        下图为超几何分布。

概率和

        注:为讨论方便起见以及主干把握要求,以下一般不再考虑定义域问题,默认为最开始讨论的情况。

        根据实际情景,我们容易得知:

        从含有10件次品的100件产品中挑选出5件产品,可能情况如下。

%5Cbegin%7Barray%7D%7B%7Cc%7Cc%7Cc%7C%7D%0A%5Chline%0A%5Ctextbf%7B%E6%96%B9%E6%A1%88%E7%BC%96%E5%8F%B7%7D%26%20%5Ctextbf%7B%E6%AD%A3%E5%93%81%E6%95%B0%7D%26%20%5Ctextbf%7B%E6%AC%A1%E5%93%81%E6%95%B0%7D%5C%5C%0A%5Chline%0A1%20%26%200%20%26%205%5C%5C%0A%5Chline%0A2%20%26%201%20%26%204%5C%5C%0A%5Chline%0A3%20%26%202%20%26%203%5C%5C%0A%5Chline%0A4%20%26%203%20%26%202%5C%5C%0A%5Chline%0A5%20%26%204%20%26%201%5C%5C%0A%5Chline%0A6%20%26%205%20%26%200%5C%5C%0A%5Chline%0A%5Cend%7Barray%7D

        因此,我们容易得到

C%5E%7B5%7D_%7B100%7D%3DC%5E%7B0%7D_%7B90%7DC%5E5_%7B10%7D%2BC%5E1_%7B90%7DC%5E4_%7B10%7D%2BC%5E2_%7B90%7DC%5E3_%7B10%7D%2BC%5E3_%7B90%7DC%5E2_%7B10%7D%2BC%5E4_%7B90%7DC%5E1_%7B10%7D%2BC%5E5_%7B90%7DC%5E0_%7B10%7D

        同样的,写成一般式就是

C%5En_N%3D%5Csum_%7Bk%3D0%7D%5E%7Bn%7D%20C%5Ek_%7BM%7DC%5E%7Bn-k%7D_%7BN-M%7D%20

        从而我们得到

%5Cbegin%7Balign%7D%0A%5Csum_%7Bk%3D0%7D%5En%20P_%7B(x%3Dk)%7D%26%3D%5Csum_%7Bk%3D0%7D%5En%20%5Cfrac%7BC%5Ek_M%20C%5E%7Bn-k%7D_%7BN-M%7D%7D%7BC_N%5En%7D%5C%5C%0A%26%3D%5Cfrac%7B%5Csum_%7Bk%3D0%7D%5En%20C%5Ek_M%20C%5E%7Bn-k%7D_%7BN-M%7D%7D%7BC_N%5En%7D%5C%5C%0A%26%3D%5Cfrac%7BC%5En_N%7D%7BC%5En_N%7D%5C%5C%0A%26%3D1%0A%5Cend%7Balign%7D

期望

        根据定义,易得

%5Cbegin%7Balign%7D%0AE(x)%26%3D%5Csum_%7Bk%3D0%7D%5En%20%5Cfrac%7B%5Ccolor%7Bblue%7D%7BkC%5Ek_M%7DC%5E%7Bn-k%7D_%7BN-M%7D%7D%7BC%5En_N%7D%5C%5C%0A%26%3D%5Csum_%7Bk%3D%5Ccolor%7Bred%7D%7B1%7D%7D%5En%20%5Cfrac%7B%5Ccolor%7Bblue%7D%7BMC%5E%7Bk-1%7D_%7BM-1%7D%7DC%5E%7Bn-k%7D_%7BN-M%7D%7D%7BC%5En_N%7D%5C%5C%0A%26%3D%5Cfrac%7BM%7D%7BC%5En_N%7D%5Csum_%7Bk%3D1%7D%5En%20C%5E%7Bk-1%7D_%7BM-1%7DC%5E%7Bn-k%7D_%7BN-M%7D%5C%5C%0A%26%3D%5Cfrac%7BM%7D%7BC%5En_N%7D%5Csum_%7Bk%3D1%7D%5En%20C%5E%7Bk-1%7D_%7BM-1%7DC%5E%7B(n-1)-(k-1)%7D_%7B(N-1)-(M-1)%7D%5C%5C%0A%26%3D%5Cfrac%7BM%7D%7BC%5En_N%7D%5Csum_%7Bk'%3D0%7D%5E%7Bn-1%7D%20C%5E%7Bk'%7D_%7BM-1%7DC%5E%7B(n-1)-k'%7D_%7B(N-1)-(M-1)%7D%EF%BC%8C%5C%20def%5C%20k'%5Cequiv%20k-1%5C%5C%0A%26%3D%5Cfrac%7BM%7D%7BC%5En_N%7DC%5E%7Bn-1%7D_%7BN-1%7D%5C%5C%0A%26%3D%5Cfrac%7BM%5Cfrac%7B(N-1)!%7D%7B(n-1)!(N-n)!%7D%7D%7B%5Cfrac%7BN!%7D%7Bn!(N-n)!%7D%7D%5C%5C%0A%26%3D%5Cfrac%7BMn%7D%7BN%7D%5C%5C%0A%26%3Dnp%EF%BC%8C%5C%20def%5C%20p%5Cequiv%20%5Cfrac%7BM%7D%7BN%7D%0A%5Cend%7Balign%7D

        与二项分布的期望相同。

        蓝色标识可参见以下文档:

        思考:从独立角度来看,二项分布的期望为np是显然的。

                那么,超几何分布呢?

方差

        根据定义,易得

%5Cbegin%7Balign%7D%0AD(x)%26%3D%5Csum_%7Bk%3D0%7D%5En%20%5Cfrac%7Bk%5Ccdot%5Ccolor%7Bblue%7D%7BkC%5Ek_M%7DC%5E%7Bn-k%7D_%7BN-M%7D%7D%7BC%5En_N%7D-E%5E2(x)%5C%5C%0A%26%3D%5Csum_%7Bk%3D%5Ccolor%7Bred%7D%7B1%7D%7D%5En%20%5Cfrac%7Bk%5Ccdot%5Ccolor%7Bblue%7D%7BMC%5E%7Bk-1%7D_%7BM-1%7D%7DC%5E%7Bn-k%7D_%7BN-M%7D%7D%7BC%5En_N%7D-E%5E2(x)%5C%5C%0A%26%3D%5Cfrac%7BM%7D%7BC%5En_N%7D%5Csum%5En_%7Bk%3D%5Ccolor%7Bred%7D%7B1%7D%7D%5Ccolor%7Bblue%7D%7B(k-1)C%5E%7Bk-1%7D_%7BM-1%7D%7DC%5E%7Bn-k%7D_%7BN-M%7D%2B%5Cfrac%7BM%7D%7BC%5En_N%7D%5Csum%5En_%7Bk%3D1%7DC%5E%7Bk-1%7D_%7BM-1%7DC%5E%7Bn-k%7D_%7BN-M%7D-E%5E2(x)%5C%5C%0A%26%3D%5Cfrac%7BM%7D%7BC%5En_N%7D%5Csum%5En_%7Bk%3D%5Ccolor%7Bred%7D2%7D%5Ccolor%7Bblue%7D%7B(M-1)C%5E%7Bk-2%7D_%7BM-2%7D%7DC%5E%7Bn-k%7D_%7BN-M%7D%2B%5Cfrac%7BMn%7D%7BN%7D-E%5E2(x)%5C%5C%0A%26%3D%5Cfrac%7BM(M-1)%7D%7BC%5En_N%7D%5Csum%5En_%7Bk%3D2%7DC%5E%7Bk-2%7D_%7BM-2%7DC%5E%7B(n-2)-(k-2)%7D_%7B(N-2)-(M-2)%7D%2B%5Cfrac%7BMn%7D%7BN%7D-E%5E2(x)%5C%5C%0A%26%3D%5Cfrac%7BM(M-1)%7D%7BC%5En_N%7D%5Csum%5E%7Bn-2%7D_%7Bk'%3D0%7DC%5E%7Bk'%7D_%7BM-2%7DC%5E%7B(n-2)-k'%7D_%7B(N-2)-(M-2)%7D%2B%5Cfrac%7BMn%7D%7BN%7D-E%5E2(x)%EF%BC%8C%5C%20def%5C%20k'%5Cequiv%20k-2%5C%5C%0A%26%3D%5Cfrac%7BM(M-1)%7D%7BC%5En_N%7DC%5E%7Bn-2%7D_%7BN-2%7D%2B%5Cfrac%7BMn%7D%7BN%7D-E%5E2(x)%5C%5C%0A%26%3D%5Cfrac%7BM(M-1)n(n-1)%7D%7BN(N-1)%7D%2B%5Cfrac%7BMn%7D%7BN%7D-%5Cfrac%7BM%5E2n%5E2%7D%7BN%5E2%7D%5C%5C%0A%26%3Dnp%5Cfrac%7B(M-1)(n-1)%7D%7BN-1%7D%2Bnp-np%5Cfrac%7BMn%7D%7BN%7D%EF%BC%8C%5C%20def%5C%20p%5Cequiv%20%5Cfrac%7BM%7D%7BN%7D%5C%5C%0A%26%3Dnp%5Cfrac%7BMn-M-n%2B1%2BN-1-(N-1)n%5Cfrac%7BM%7D%7BN%7D%7D%7BN-1%7D%5C%5C%0A%26%3Dnp%5Cfrac%7BN%5E2-NM-Nm%2BMn%7D%7BN(N-1)%7D%5C%5C%0A%26%3D%5Cfrac%7BMn(N-M)(N-n)%7D%7BN%5E2(N-1)%7D%5C%5C%0A%26%3Dnp(1-p)%5Cfrac%7BN-n%7D%7BN-1%7D%0A%5Cend%7Balign%7D

        推导思路其实大同小异。

        与二项分布的方差D(x)%3Dnp(1-p)相比,多出了最后一个因子。

        易知当n%3E1时,超几何分布的方差比二项分布的方差小。

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